What is the arc length of #f(x) = -cscx # on #x in [pi/12,(pi)/8] #?
1 Answer
Explanation:
#f(x)=-cscx#
#f'(x)=cscxcotx#
Arc length is given by:
#L=int_(pi/12)^(pi/8)sqrt(1+csc^2xcot^2x)dx#
Rearrange:
#L=int_(pi/12)^(pi/8)cscxcotxsqrt(1+sin^2xtan^2x)dx#
For
#L=int_(pi/12)^(pi/8)cscxcotx{sum_(n=0)^oo((1/2),(n))(sinxtanx)^(2n)}dx#
Isolate the
#L=int_(pi/12)^(pi/8)cscxcotxdx+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)(sinxtanx)^(2n-1)dx#
Rearrange:
#L=[-cscx]_ (pi/12)^(pi/8)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)(secx-1/secx)^(2n-1)dx#
Apply binominal expansion:
#L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)sum_(k=0)^(2n-1)((2n-1),(k))(secx)^(2n-1-k)(-1/secx)^kdx#
Rearrange:
#L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oosum_(k=0)^(2n-1)((1/2),(n))((2n-1),(k))(-1)^kint_(pi/12)^(pi/8)sec^(2(n-k)-1)xdx#